Radicals, Radicals, And More Radicals in an Inequality

Problem

Solution

If $x=z,\,$ then $x=y=z\,$ and there is nothing to proof. So assume $x\lt z.\,$ Introduce $t=\displaystyle\sqrt{\frac{y}{x}}\,$ and $k=\displaystyle\sqrt{\frac{z}{x}}.\,$ Obviously, $1\le t\le k,\,$ and we have to prove that

The only critical point of $f\,$ is $\displaystyle t'=\frac{1+k^2}{1+k}\,$ and $1\lt t'\lt k,\,$ implying that $\min f\in \{f(1),f(k)\}.\,$ But $f(1)\gt 1\,$ and $f(k)\gt 1,\,$ which completes the proof.

Acknowledgment

The unusual problem by Lorian Saceanu has been kindly communicated to me by Leo Giugiuc, along with his solution.

Three Junior Problems from Vietnam $\left(\left(\begin{array}{ccc}0&y&z\\x&0&z\\x&y&0\end{array}\right)\left(\begin{array}\;a\\b\\c\end{array}\right)=\left(\begin{array}\;x\\y\\z\end{array}\right)\right)$

Four Integrals in One Inequality $\left(\displaystyle \small{\left(\int_a^bxf(x)dx\right)\left(\int_a^bf^2(x)dx\right)\left(\int_a^bx^3f(x)dx\right)\ge\frac{a^2b^2}{b-a}\left(\int_a^bf(x)dx\right)^4}\right)$